The Matrix Transformations on Orlicz Space of Χπ

Applied Mathematical Sciences, Vol. 8, 2014, no. 178, 8889 - 8896 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2014.49741

The Matrix Transformations on Orlicz Space of χπ

B. Baskaran and S. Thalapathiraj

Department of Mathematics SRM University, Vadapalani Campus Chennai-26, Tamilnadu, India

Copyright c 2014 B. Baskaran and S. Thalapathiraj. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Let χπ denote the space of all χπ sequences and Λ the space of all analytic rate sequences. First we show that the set E = {s(k) : k = π 1, 2, 3, ..} is a determining set for χM . The set of all ﬁnite matrices π π transforming χM in to FK- space Y denoted by (χM : Y ) . We charac- π π terize the classes (χM : Y ) , When Y = (c0)π, cπ, χM , `π, `s, Λπ, hπ. In summary we have the following table:

π % (c0)π cπ χM `π `s Λπ hπ π χM Necessary and suﬃcient condition on the matrix are obtained

But the approach to obtain these result in the present paper is by π π determining set for χM . First , we investigate a determining set for χM and then we characterize the classes of matrix transformations involving π χM and other known sequence spaces.

Mathematics Subject Classiﬁcation: 40A05, 40C05, 40D05

Keywords: Orlicz space,Analytical sequence,rate sequence,entire sequence 8890 B. Baskaran and S. Thalapathiraj

1 Introduction

th A complex sequence, whose k terms is xk is denoted by {xk} or simply x. Let ω be the set of all sequences x = {xk} and φ be the set of all ﬁnite sequences. Let `∞, c, c0 be the sequence spaces of bounded, convergent and null sup sequences x = {xk} respectively. In respect of `∞, c, c0 we have kxk =k |xk|, where x = {xk} ∈ c0 ⊂ c ⊂ `∞. A sequences x = {xk} is said to be analytic if sup 1 k |xk| k < ∞ . The vector space of all analytic sequences will be denoted by lim 1 Λ. A sequence x is called entire sequences if k→∞|xk| k = 0. The vector space of all entire sequences will be denoted by Γ. χ was discussed in Kamthan [11]. Matrix transformations involving χ were characterized by sridhar [12] and Sirajudeen [13] . Let χ be the set of all those sequences x = {xk} . such 1 that (k!|xk|) k → 0 as k → ∞ . Then χ is a metric space with the metric sup 1 d(x, y) =k {(k!|xk − yk|) k : k = 1, 2, 3, ...... }

Orlicz [17] used the idea of Orlicz function to construct the space (LM ). Lindenstrauss and Tzafriri [1] investigated Orlicz sequence spaces in more de- tail, and they proved that every Orlicz sequence space `M contains a subspace isomorphic to `p (1 ≤ p < ∞) . Subsequently diﬀerent classes of sequence spaces were deﬁned by Parashar and Choudhry [2] , Mursaleen [3] , Bektas and Altin [4] , Tripathy .[5], Rao and subramanian[6],and many others. The Orlicz sequence spaces are the special cases of Orlicz spaces studied in [7]. An Orlicz function is a function M : [0, ∞) → [0, ∞) which is continuous,non- decreasing and convex with M(0) =0, M(x)> 0, x> 0 and M(x) → ∞ as x → ∞. If the convexity of Orlicz function M is replaced by M(x+y)≤ M(x)+M(y) then this function is called modulus function , introduced by Nakano [17] and further discussed by Ruckle [8] and Maddox[9], and many others.

An Orlicz function M is said to satisfy 42 condition for all values of u , if there exists a constant K > 0 , such that M(2u) ≤KM(u) (u ≥0). The condition 42 is equivalent to M(`u) ≤ K ` M(u),for all values u and for ` > 1.Lindestrauss and Tzafriri [1] used the idea of Orlicz function to construct Orlicz sequence space ∞ |xk| `M ={ x ∈ w :Σk=1 M( ρ )< ∞, for some ρ > 0 }...... (1)

The space `M with the norm ∞ |xk| kxk = inf{ρ >0:Σk=1 M( ρ )≤1}...... (2)

becomes a Banach space which is called an Orlicz sequence space. For M(t) p = t ,(1 ≤ p < ∞) the space `M coincide with the classical sequence space `p. th n Given sequence x={xk} its n section is the sequence x = {x1, x2, ...... , xn, 0, 0, ....} δn= (0, 0, ....., 1, 0, 0, ...... ),, 1 in the nth place and zero’s else where ; and sk= The matrix transformations on Orlicz space of χπ 8891

(0, 0, ....., 1, −1, 0, ...) ,1 in the nth place ,-1 in the (n + 1)th place and zero’s else where .. If X is a sequence space, we define (i) X 0 = the continuous dual of X α P∞ (ii)X = {a = (ak): K=1 |akxk| < ∞, for each x ∈ X} β P∞ (iii)X = {a = (ak): K=1 akxk is convergent, for each x ∈ X} γ sup P∞ (iv)X = {a = (ak):n | K=1 akxk| < ∞, for each x ∈ X} (v) Let X be an FK space ⊃ φ . Then Xf = {f(δn): f ∈ X 0 } Xα,Xβ,Xγ are called the α -(or KO¨ the -To ëplitz) dual of X , β-( or generalized Ko ¨ the - To ëplitz ) dual of X, γ- dual of X. Note that Xα⊂Xβ ⊂ Xγ. If X⊂ Y then Y µ⊂Xµ, for µ = α,β, or γ. An FK space (Frechet coordinate space ) is a Frechet space which is made up of numerical sequences and has the property that the coordinate functionals pk(x) = xk(k= 1,2,3,....) are continuous . An FK - space is a locally convex Frechet space which is made up of sequences and has the property that coordinate projections are continuous. An metric space (X,d) is said to have AK (or sectional convergence) if and only if d(xn, x)→ x as n→ ∞.[10] The space is said to have AD (or) be an AD space if ∅ is dense in X .We note that implies AK implies AD by [4].

Lemma 1.1 Let X be an FK-space ⊃ φ. Then (i) Xγ ⊂ Xf (ii)If X has AK, Xβ = Xf (iii) If X has AD, Xβ = Xγ.

2 Deﬁnition and Preliminaries

∞ Deﬁnition 2.1 Let ω denote the set of all complex double sequence x = {xk}k=1 and M : [0, ∞) → [0, ∞) be an Orlicz function, or a modulus function. Deﬁne the sets 1 π lim (k!|xk|) k χM = {x ∈ ω : k→∞(M( 1 )) = 0 for some ρ > o} (πk) k ρ 1 lim (|xk|) k ΓM = {x ∈ ω : k→∞(M( 1 )) = 0 for some ρ > o} (πk) k ρ 1 sup (|xk|) k ΛM = {x ∈ ω : k (M( 1 )) < ∞ for some ρ > o} (πk) k ρ π The space χM is a metric space with the metric 1 sup (k!|xk−yk|) k d(x,y) = inf {ρ > o : k (M( 1 )) ≤ 1} (πk) k ρ π π The space ΓM and ΛM is a metric space with the metric 8892 B. Baskaran and S. Thalapathiraj

1 sup (|xk−yk|) k d(x,y) = inf {ρ > o : k (M( 1 )) ≤ 1} (πk) k ρ Let `s denotes the space of all those complex sequences x = {xk} such that {x1, x1 + x2, x1 + x2 + x3 + .... + x1 + x2 + .... + xk + ...... } belongs to ` with the norm kxks = |x1| + |x1 + x2| + ...... + |x1 + x2 + ...... + xk| + ...... xk xk Γπ = {x = {xk} :( ) ∈ Γ)} and Λπ = {x = {xk} :( ) ∈ Λ} πk πk

Then Γπ and Λπ are FK-spaces with the metric 1 sup xk−yk xk d(x,y) = {| | k : k = 1, 2, 3, ...... }. hπ = {x = {xk} :( ) ∈ h} Then k πk πk hπ is a BK-space with the norm kxk = Σ∞ k| xk − xk+1 |. k=1 πk πk+1 xk (`∞)π = mπ = {x = {xk} ∈ ω :( ) ∈ m} πk xk xk (c0)π = {x = {xk} ∈ ω :( ) ∈ c0} , (c)π = {x = {xk} ∈ ω :( ) ∈ c}. πk πk sup xk In respect of mπ, (c0)π, cπ are BK-spaces with the norm kxkπ = | |, k πk xk (`)π = {x = {xk} ∈ ω :( ) ∈ `} , πk ∞ xk `π is a BK-space with the norm kxk = Σ | |. k=1 πk we call (c0)π, cπ, `π, Λπ, hπ are rate spaces. Let X be an BK-space. Then D = D(X) = {x ∈ φ : kxk ≤ 1} we do not assume that X ⊃ φ (i.e) D = φ T(unit closed sphere in X) Let X be an BK space. A subset E of φ will be called a determining set for X if D(X) is the absolutely convex hull of E. In respect of a metric space (X,d), D= {x ∈ φ : d(x, 0) ≤ 1} Given a sequence x = {xk} and an inﬁnite matrix A = (ank), (n, k = 1, 2, ....) ∞ then A - transform of x is the sequence y = (yn) where yn = Σk=1ankxk(n, k = 1, 2, 3, ...... ), whenever Σankxk exists. Let X and Y be FK- spaces. If y ∈ Y whenever x ∈ X ,then the class of all matrices A is denoted by (X : Y )

Lemma 2.2 Let X be an FK- space and E is determining set for X. Let Y be an FK-space and A is a inﬁnite matrix. Suppose that either X has AK or A is row ﬁnite.Then A ∈ (X : Y ) if and only if (1) The columns of A belong to Y and (2) A[E] is a bounded subset of Y.

3 Main Results

π Proposition 3.1 χM has AK,where M is a modulus function. 1 π (k!|xk|) k π Proof: Let x = {xk} ∈ χM , then {M( 1 )} ∈ χ and hence (πk) k ρ 1 sup (k!|xk|) k k≥n+1((M( 1 )) → 0 as n → ∞ (πk) k ρ Therefore The matrix transformations on Orlicz space of χπ 8893

1 [n] sup (k!|xk|) k d(x, x ) = inf {ρ > o : k≥n+1(M( 1 )) ≤ 1} → 0 as n → ∞ (πk) k ρ [n] π ⇒ x → x as n → ∞, implying that χM has AK. This completes the proof.

Proposition 3.2 Let {s(k) : k = 1, 2, 3, ...} be the set of all sequences in φ each of whose nonzero terms ±1 .Let E = {s(k) : k = 1, 2, 3, ...... } then E is a π determining set for the space χM . π Proof:Step 1: Recall that χM is a metric space with the metric 1 sup (k!|xk−yk|) k d(x,y) = inf {ρ > o : k (M( 1 )) ≤ 1} (πk) k ρ Let A be the absolutely convex hull of E . m (k) m (k) Let x ∈ A .Then x = Σk=1tks with Σk=1|tk| ≤ 1 and s ∈ E ...(a) (1) (2) (m) Then d(x, 0) ≤ |t1|d(s , 0) + |t2|d(s , 0) + ...... + |tm|d(s , 0). (k) m But d(s , 0) = 1 for k =1,2,3,.....,m.Hence d(x, 0) ≤ Σk=1|tk| ≤ 1 by using (a) Also x ∈ φ,Hence x ∈ D.Thus we have A ⊂ D Step 2: Let x ∈ D ⇒ x ∈ φ and d(x, 0) ≤ 1 ⇒ x = {x1, x2, ...... , xm} and 1 1 1 (1!|x1|) 1 (2!|x2|) 2 (m!|xm|) m sup{M( 1 ),M( 1 ), ...... ,M( 1 )} ≤ 1....(b) (πk) k ρ (πk) k ρ (πk) k ρ 1 1 1 (1!|x1|) 1 (2!|x2|) 2 (m!|xm|) m case(i): suppose that M( 1 ) ≥ M( 1 ) ≥ ...... ≥ M( 1 ) (πk) k ρ (πk) k ρ (πk) k ρ

(i!|xi|) M( 1 ) (i!xi) (πk) k ρ Let ∈i= sgn(M( 1 )) = (i!xi) for i = 1, 2, 3, ....., m (πk) k ρ M( 1 ) (πk) k ρ Take Sj = { ∈2, ∈2, ...... , ∈j, 0, 0, ...... } for j = 1,2,3,...... ,m. 1 1 (1!|x1|) 1 (2!|x2|) 2 Then Sj ∈ E for j = 1, 2, 3, ....., m Also x = (M( 1 ) − M( 1 ))s1 (πk) k ρ (πk) k ρ 1 1 1 1 m+1 (2!|x2|) 2 (3!|x3|) 3 (m!|xm|) m ((m+1)!|x(m+1)|) +M( 1 )) − M( 1 ))s2 + ...... +(M( 1 ) − M( 1 )sm (πk) k ρ (πk) k ρ (πk) k ρ (πk) k ρ = t1s1 + t2s2 + ..... + tmsm.so that 1 1 1 m+1 (1!|x1|) 1 ((m+1)!|x(m+1)|) (1!|x1|) 1 t1 + t2 + ..... + tm = (M( 1 ) − M( 1 )= M( 1 ) (πk) k ρ (πk) k ρ (πk) k ρ 1 m+1 ((m+1)!|x(m+1)|) because M( 1 ) = 0 (πk) k ρ Therefore t1 + t2 + ...... + tm ≤ 1 by using (b). Hence x ∈ A.Thus we have D ⊂ A.

Case(ii): Let y be x and let (1!|y1|) (2!|y2|) (m!|ym|) M( 1 ) ≥ M( 1 ) ≥ ...... ≥ M( 1 ) Express y as a member of A as (πk) k ρ (πk) k ρ (πk) k ρ in case(i). Since E is invariant under permutation of the terms of its members, so is A. 8894 B. Baskaran and S. Thalapathiraj

Hence x ∈ A. Thus D ⊂ A. Therefore in both cases D ⊂ A ...(c) From (a) π and (c) A=D. Consequently E is a determining set for χM . This completes the proof.

Proposition 3.3 An inﬁnite matrix A = (ank) is in the class

A ∈ (χπ :(c ) ) ⇔lim ( ank ) = 0...(d) M 0 π n→∞ πn ⇔sup | an1+.....+ank | < ∞....(e) nk πn Proof: π In lemma 2.3.Take X = χM has AK property take Y = (c0)π be an FK-space. π Further more χM is a determining set E (as in given Proposition 2.6). Also (k) π A[E] = A(s ) = {(an1 + an2 + .....)}. Again by lemma2.3.A ∈ (χM :(c0)π) if and only if (i) The columns of A belongs (c0)π and (k) (ii) A(S ) is a bounded subset (c0)π.But the condition (i) ⇔ { ank : n = 1, 2, ....} is exits for all k. πn (ii) ⇔sup | an1+an2+....+ank | < ∞ n, k πn π Hence we conclude that A ∈ (χM :(c0)π) ⇔ conditions (d) and (e) are satis- ﬁed. This completes the proof.Omitting this proofs ,we formulate the following results:

Proposition 3.4 An inﬁnite matrix A = (ank) is in the class A ∈ (χπ :(c )) ⇔ lim ( ank ) exists (k = 1, 2, 3, ....)...(f) M π n→∞ πn ⇔sup | an1+.....+ank | < ∞ nk πn

Proposition 3.5 An inﬁnite matrix A = (ank) is in the class 1 π π sup (n!|an1+.....+ank| n ) A ∈ (χM : χM ) ⇔ nk (M( 1 )) < ∞ (πk) k ρ 1 lim (n!|ank|) n ⇔ n→∞(M( 1 )) = 0, for(k = 1, 2, 3, ....) (πk) k ρ π ⇔ d(an1, an2, ...., ank) is bounded for each metric d on χM and for all n, k

Proposition 3.6 An inﬁnite matrix A = (ank) is in the class π ∞ A ∈ (χM : `π) ⇔ Σn=1|ank| converges (k = 1, 2, 3, ...) ⇔sup Σ∞ | ank | < ∞ nk n=1 πn

π Proposition 3.7 An inﬁnite matrix A = (ank) is in the class A ∈ (χM : `s) sup ∞ ⇔ k Σn=1|ank| < ∞

Proposition 3.8 An inﬁnite matrix A = (ank) is in the class 1 π sup k anγ A ∈ (χ :Λ ) ⇔ (Σ | | n ) < ∞ M π n k γ=1 πn ⇔ d(an1, an2, ....., ank) is bounded for each metric d on Λπ and for all n, k The matrix transformations on Orlicz space of χπ 8895

π Proposition 3.9 An inﬁnite matrix A = (ank) is in the class A ∈ (χM : hπ) ⇔ { ank : n = 1, 2, 3, ...} is exists for each k πn a +a +.....+a ⇔ supΣ∞ | an1+an2+....+ank − n+1,1 n+2,2 n+k,k | < ∞ k n=1 πn πn+1 Acknowledgements. The second author dedicate this journal to Dr. N. Subramanian, Professor, Department of Mathematics, Sastra University, Thanjavur, Tamil Nadu, India.

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Received: September 21, 2014; Published: December 12, 2014